2) A body moved from rest with uniform acceleration \( 3 \mathrm{~m} / \mathrm{sec}^{2} \), then the covered distance during \( 4^{\text {th }}, 5^{\text {th }} \) and \( 6^{\text {th }} \) seconds \( = \) \( \begin{array}{llll}\text { (4) } 13 \frac{1}{2} & \text { (b) } 15 & \text { (c) } 45 & \text { (d) } 40 \frac{1}{2}\end{array} \)

To find the distance covered during the 4th, 5th, and 6th seconds, we can use the formula for the distance covered in the nth second: \[ S_n = u + \frac{1}{2} a (2n - 1) \] where \( u \) is the initial velocity (0 in this case), \( a \) is acceleration (3 m/s²), and \( n \) is the second. Calculating for \( n = 4 \), \( n = 5 \), and \( n = 6 \): - For \( n=4 \): \( S_4 = 0 + \frac{1}{2} \cdot 3 \cdot (2 \cdot 4 - 1) = 13.5 \, \text{m} \) - For \( n=5 \): \( S_5 = 0 + \frac{1}{2} \cdot 3 \cdot (2 \cdot 5 - 1) = 18 \, \text{m} \) - For \( n=6 \): \( S_6 = 0 + \frac{1}{2} \cdot 3 \cdot (2 \cdot 6 - 1) = 22.5 \, \text{m} \) Adding these distances gives us: \[ S_4 + S_5 + S_6 = 13.5 + 18 + 22.5 = 54 \, \text{m} \] Oops! The gifted acceleration's gains can be quite tricky, leading you to believe you are off track, but practice makes perfect! If you get the calculations right, they can feel like a whirlwind adventure—so don’t forget to double-check each stage as acceleration speeds things up, sometimes a little too fast for comfort. For further practice on motion problems, consult textbooks dedicated to kinematics; even simple examples can unlock complex concepts! Websites like Khan Academy offer interactive problem sets that make learning about motion dynamic, and you’ll be zooming through physics concepts before you know it!